Using the Normal Distribution Revision Notes for HSC SSCE Mathematics Advanced

Using the Normal Distribution

Introduction to real-world applications

The normal distribution appears throughout everyday life and practical situations. Many natural phenomena and manufactured processes follow a normal distribution pattern, or approximate it closely enough to make calculations useful and reliable. This makes the normal distribution one of the most important tools in statistics and probability.

Note

When working with normal distribution problems, you can use different approaches depending on what information you need. The empirical rule provides quick estimates for common situations, while standard normal probability tables give more precise values for any scenario.

Understanding the empirical rule

The empirical rule, also called the 68-95-99.7 rule, is a quick method for estimating probabilities when values fall within 1, 2, or 3 standard deviations of the mean. This rule applies to all normal distributions and provides these key percentages:

Approximately 68% of values fall within 1 standard deviation of the mean: $P(-1 \leq Z \leq 1) = 68\%$
Approximately 95% of values fall within 2 standard deviations of the mean: $P(-2 \leq Z \leq 2) = 95\%$
Approximately 99.7% of values fall within 3 standard deviations of the mean: $P(-3 \leq Z \leq 3) = 99.7\%$

Important

The empirical rule works particularly well for quick mental calculations and initial estimates. When you need more precision, or when dealing with values that don't fall exactly on 1, 2, or 3 standard deviations, you should use standard normal probability tables or technology.

Calculating with standard deviations

To use the normal distribution effectively, you need to determine how many standard deviations a particular value sits from the mean. This measurement is called a z-score or standardized score.

Calculate the z-score by finding the difference between your value and the mean, then dividing by the standard deviation. For example, if the mean is 102 and a value of 100 has a standard deviation of 2, then 100 sits 1 standard deviation below the mean because $(100 - 102) \div 2 = -1$ .

Note

The negative sign indicates the value falls below the mean, while a positive z-score means the value exceeds the mean.

Using complementary probabilities and symmetry

Two important properties help simplify normal distribution calculations: the complement rule and symmetry.

The complement rule states that probabilities must sum to 1 (or 100%). If you know that 68% of values fall between $Z = -1$ and $Z = 1$ , then the remaining 32% must fall outside this range. This gives us:

$P(Z < -1 \text{ or } Z > 1) = 100\% - 68\% = 32\%$

The normal distribution curve is perfectly symmetric around its mean. This symmetry means equal areas appear on both sides of the centre. Therefore, if 32% falls outside the range $-1$ to $1$ , exactly half (16%) falls below $Z = -1$ and half (16%) falls above $Z = 1$ :

$P(Z < -1) = 32\% \div 2 = 16\%$

Reading standard normal tables

Standard normal probability tables show cumulative probabilities for different z-scores. These tables tell you the probability that a value falls below a particular z-score.

When using these tables, read the z-score you want to find (usually to one decimal place) and the table provides the corresponding probability. For example, if the table shows $P(Z \leq 1.5) = 93\%$ , this means 93% of values fall at or below 1.5 standard deviations above the mean.

Note

To find the probability above a certain z-score, use the complement rule. Since probabilities sum to 100%, subtract the table value from 100%:

$P(Z > 1.5) = 100\% - 93\% = 7\%$

Worked example: chocolate bar manufacturing

Consider a chocolate company that manufactures 100 g chocolate-nougat bars. Manufacturing processes naturally create small variations, so not every bar weighs exactly 100 g. The company knows these weights follow a normal distribution with a standard deviation of 2 g. To reduce customer complaints, the company adjusts its machinery to produce bars with a mean weight of 102 g.

Example

Worked Example Part A(i): Finding the percentage below stated weight

We need to determine what percentage of chocolate bars weigh less than the stated 100 g.

Step 1: Calculate the z-score

The mean weight is 102 g, so 100 g is 2 g below the mean. With a standard deviation of 2 g, this means 100 g sits exactly 1 standard deviation below the mean ( $Z = -1$ ).

Step 2: Apply the empirical rule

$P(-1 \leq Z \leq 1) = 68\%$

This tells us 68% of bars fall between 1 standard deviation below and 1 standard deviation above the mean.

Step 3: Use the complement rule

$P(Z < -1 \text{ or } Z > 1) = 100\% - 68\% = 32\%$

Step 4: Apply symmetry

By symmetry, half of this 32% falls below $Z = -1$ :

$P(Z < -1) = 32\% \div 2 = 16\%$

Answer: 16% of chocolate bars weigh less than the stated 100 g.

Example

Worked Example Part A(ii): Finding the percentage above a higher weight

We need to find what percentage of bars weigh more than 105 g.

Step 1: Calculate the z-score

105 g is 3 g above the mean of 102 g. With a standard deviation of 2 g, this gives:

$Z = (105 - 102) \div 2 = 1.5$

Step 2: Consult the standard normal table

The value 1.5 doesn't match the standard values (1, 2, or 3) used in the empirical rule, so we need to consult a standard normal table. The table shows:

$P(Z \leq 1.5) = 93\%$

This means 93% of bars weigh 105 g or less.

Step 3: Use the complement rule

$P(Z > 1.5) = 100\% - 93\% = 7\%$

Answer: 7% of chocolate bars weigh more than 105 g.

Example

Worked Example Part B: Working backwards from a probability target

Suppose the company wants fewer than 1 in 1000 bars to weigh under 100 g. What should the mean weight be?

Step 1: Identify the target probability

We want $P(Z < z_0) = 0.001$ where $z_0$ represents the z-score for 100 g.

Step 2: Read the table backwards

Reading the standard normal table backwards, we find:

$P(Z \leq 3.1) = 0.999$

Step 3: Apply symmetry and complements

$P(Z \leq -3.1) = 1 - 0.999 = 0.001$

This tells us that 100 g needs to be 3.1 standard deviations below the new mean.

Step 4: Calculate the required mean

With a standard deviation of 2 g:

$\mu = 100 + 3.1 \times 2 = 106.2 \text{ g}$

Answer: The company should set the mean weight to 106.2 g to achieve their quality target.

Practical strategies for solving problems

When approaching normal distribution problems, follow this systematic approach:

Start by clearly identifying the mean ( $\mu$ ) and standard deviation ( $\sigma$ ) of your distribution. Then determine what value you're interested in and calculate how many standard deviations it sits from the mean.

Note

Key Decision Point:

If your value falls exactly 1, 2, or 3 standard deviations from the mean, use the empirical rule for a quick solution. Otherwise, consult standard normal probability tables or use technology.

Remember to use the complement rule when you need the probability above a certain value, as tables typically show cumulative probabilities (below the value). Take advantage of symmetry when working with values below the mean.

Important

For problems asking you to find a required mean or standard deviation, work backwards from the probability using tables to find the z-score, then convert this back to the original units.

Remember!

Summary

Key Points to Remember:

The normal distribution models many real-world situations, making it extremely useful for practical calculations
The empirical rule provides quick estimates: approximately 68%, 95%, and 99.7% of values fall within 1, 2, and 3 standard deviations of the mean
Calculate z-scores to determine how many standard deviations a value sits from the mean
Use complement rule: probabilities outside a range equal 100% minus the probability inside the range
The normal curve's symmetry means equal areas appear on both sides of the mean
Standard normal tables show cumulative probabilities, so use complements to find probabilities above a value
Working backwards from probability to find required parameters is a key skill in quality control and practical applications

Using the Normal Distribution (HSC SSCE Mathematics Advanced): Revision Notes